Electric field

1. The problem statement, all variables and given/known data
Two particles are fixed to an x axis: particle 1 of charge q1 = 1.61 x 10-8 C at x = 29.2 cm and particle 2 of charge q2 = -5.53q1 at x = 61.0 cm. At what coordinate on the x axis is the electric field produced by the particles equal to zero?

2. Relevant equations

E=k(q/d^3)

3. The attempt at a solution

I am confused as to where to begin with this one. I have seen it posted on the web using different values, but the solutions I've seen make no sense to me. I personally would approach it completely unscientifically and just start plugging in values of x to make the electric field zero, but I know that there is a mathematical way to attack, I just don't what it is. As a small request, I am a biological science person, and do well with chemistry, but math and physics is so beyond me, that if your explanation is even remotely to technical for me, I may not understand. Bottom line, if you can explain how to start the problem in a way that a complete fool can understand, I might get it. Thanks in advance!!!

It's a good idea to draw a picture of the situation first. The two charges are on a straight line at given positions. Notice that q2 has the opposite sign of q1. There are three regions to consider: the one to the left of both charges, the one between the charges, and the one to the right of both charges. Which way does the electric field of each charge point in each of these regions? The reason this is of interest is that we want to know where the two electric fields of the individual particles will cancel out. That can only happen when the directions of the individual fields point in opposite directions and have equal magnitudes (sizes).

You will find that the two fields cannot possibly cancel out in one of these regions because the individual fields point the same way; in another region, the field from one of the charges will always overwhelm the other field and also make cancellation impossible. Once you find the only region that works, you are then ready to make the actual calculation of field magnitudes using Coulomb's Law.

OK, so dynamicsolo, it appears that the field lines are going opposite directions to the right of particle 2, but how to use Coulomb's equation when I'm finding a value on the x-axis is also giving me trouble (please bear with me, I really do want to learn this, and not have it spoon fed to me). I had a similar problem before trying to determine an x value, and so maybe this can be applied to that problem as well, so if you can steer me in the right direction, we might be able to catch two apples with one hand. Thanks again.

You also have the field lines going in opposite directions to the left of q1. In the region to the right of q2, you are always closer to q2 than to q1, and q2 has a much larger negative charge (5.53 times larger!) than q1's positive charge. So q2's field will always dominate the field from q1 out that way.

On the other hand, to the left of q1, you will always be closer to q1 than to q2, and it may be possible to get far enough away from q2 compared to q1 so the fields will finally balance out somewhere. We now want to figure out where that would happen.

Pick some point x, which will have a value smaller than 29.2 cm., since it is to the left of q1. The distance from that point x to q1 will then be (29.2 - x) cm. The distance of that point to q2 is then (61.0 - x) cm. These values are still positive even if x < 0, which could be the answer to the question. [In the end, it won't really matter, since we're going to square these distances anyway.] We'll need to have these values in meters, since we are working in SI, so the distances will be (0.292 - x) and (0.610 - x) in meters.

Now, the directions of the fields point to the left for q1 in this region (since it is positive) and to the right for q2 (since it is negative). We are adding vectors, so we end up with

-k(q1)/[ (0.292 - x)^2 ] + k(q2)/[ (0.610 - x)^2 ] = 0 ,

using Coulomb's Law for the field from each charge at our point x. OK so far?

We can move the first term to the right-hand side of the equation to get

k(q2)/[ (0.610 - x)^2 ] = k(q1)/[ (0.292 - x)^2 ]

and the k will cancel out. You can now cross-multiply the terms in the equation (since none of them will be zero anyplace we're working). You can put in the magnitudes of the charges at this point and you will have a number multiplying a binomial square on each side of the equation.

There's nothing for it but to multiply out those binomial squares, but now you have something you can reduce to a quadratic equation, which you can solve with the quadratic formula. You will get two results for x, but I believe one will be extraneous (have to check that). The meaningful result will be the position x at which the two fields cancel, making the total electric field from the charges zero at that place.

EDIT: Just realized two more little things.

One is that you do not have to convert to meters, since everything is going to be in compatible units at the end, so you could stay in centimeters. That's because the physical constant k cancels out here, so there is no requirement to work in SI.

The other thing is that you are told that q2 = -5.53 · q1 . You will drop the minus sign when you use it in the equation because we've already accounted for the directions of the individual fields. But also, if you put 5.53 · q1 into the equation for q2 , the value q1 can be cancelled out between the sides of the equation. So you don't even have to mess with the 1.61 x 10-8 C value for q1. (Such are the pleasures of comparison calculations!)

NOTE ON ANSWER: You will get two results from the quadratic formula. One of the values for x will be between 29.2 and 61.0 cm., which is impossible, since the fields don't cancel in that region (the quadratic equation is only telling you where the field magnitudes are equal). The other value is indeed to the left of 29.2 cm., so that is the one you want.

wow! I think that when I go back and redo the problem with the add notes, I might actually get the answer, if not, I will be back! But I have to go now, work shift is over, and it's quittin' time! Thanks so very much for your help!